Question

Tìm x, y, z biết:
\(\dfrac{1}{x-1}=\dfrac{2}{y-2}=\dfrac{3}{z-3}\) và \(x+2y+3z=56\)

Answer 1

Giải:
Ta có: \(\dfrac{1}{x-1}=\dfrac{2}{y-2}=\dfrac{3}{z-3}\Leftrightarrow\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}\)

Đặt \(\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=k+1\\y=2k+2\\z=3k+3\end{matrix}\right.\)

Mà \(x+2y+3z=56\)

\(\Leftrightarrow k+1+4k+4+9k+9=56\)

\(\Leftrightarrow14k=42\)

\(\Leftrightarrow k=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=8\\z=12\end{matrix}\right.\)

Vậy bộ số \(\left(x;y;z\right)\) là \(\left(4;8;12\right)\)